Question:
The curve amongst the family of curves, represented by the differential equation, $\left(x^{2}-y^{2}\right) d x+2 x y d y=0$ which passes through $(1,1)$ is :
Correct Option: , 2
Solution:
$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$
$\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$
Put $\quad y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
Solving we get,
$\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}$
$\ln \left(v^{2}+1\right)=-\ln x+C$
$\left(y^{2}+x^{2}\right)=C x$
$1+1=C \Rightarrow C=2$
$y^{2}+x^{2}=2 x$
$\therefore$ Option (2)